Determine the limiting reactant for this reaction if 3.027 grams of Aluminum are reacted with 60.00

mL of a solution that is 6 M KOH. Determine the amount of KAl(SO4)2•12 H2O that can be produced from this reaction. Molar mass Alum = 474 g/mol

2 Al (s) + 2 KOH (aq) + 22 H2O (l) + 4 H2SO4 (aq) → 2 KAl(SO4)2 •12 H2O (s) + 3 H2 (g)

mols Al = grams/molar mass = ?

mols KOH = M x L = ?

Use the coefficients in the balanced equation to convert mols Al to mols of the product.
Do the same for the mols KOH to mols of the product.
It is likely that the two values for mols product will not agree; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that result is the limiting reagent.

Use the smaller value and convert to grams alum. g = mols x molar mass = ?

To determine the limiting reactant, we need to calculate the number of moles of each reactant first.

1. Calculate the number of moles of Aluminum (Al):
Number of moles = mass / molar mass
Number of moles of Al = 3.027 g / 27 g/mol (molar mass of Al) = 0.112 moles

2. Calculate the number of moles of KOH:
Number of moles = concentration (M) × volume (L)
Volume of KOH = 60.00 mL = 0.06000 L
Number of moles of KOH = 6 M × 0.06000 L = 0.360 moles

3. Now, we compare the moles of both reactants to determine the limiting reactant. Since the stoichiometry of the balanced equation shows that 2 moles of Al react with 2 moles of KOH, we can divide the number of moles of each reactant by their respective coefficient (2 moles for both):
Moles of Al/2 = 0.112 moles / 2 = 0.056 moles of Al
Moles of KOH/2 = 0.360 moles / 2 = 0.180 moles of KOH

The smaller value is 0.056 moles of Al. Therefore, Aluminum (Al) is the limiting reactant.

To determine the amount of KAl(SO4)2 •12 H2O that can be produced from this reaction, we need to use the stoichiometry of the balanced equation.

4. From the balanced equation, we see that 2 moles of Al produces 2 moles of KAl(SO4)2 •12 H2O. Therefore, the moles of KAl(SO4)2 •12 H2O produced is equal to the moles of Al used.

Moles of KAl(SO4)2 •12 H2O = 0.056 moles

5. To determine the mass of KAl(SO4)2 •12 H2O, we can use the following equation:

Mass = moles × molar mass
Mass of KAl(SO4)2 •12 H2O = 0.056 moles × 474 g/mol (molar mass of KAl(SO4)2 •12 H2O)

You can calculate this final value to obtain the amount of KAl(SO4)2 •12 H2O that can be produced from the given reaction.